Question

An electron is accelerated through a potential of 120V. Find its de Broglie

wavelength ​

Answer 1

Given : Entity in question - electron

Potential : 120 Volt

Find : de Broglie wavelength of electron at given potential.

Solution : To find de Broglie wavelength with respect to potential, the formula used is : lambda = h/√(2*m*e*V). In the formula, h is planck's constant, m is mass of electron, e is charge of electron and V is potential difference.

Solving by keeping the constant values, the formula will be :

lambda = 1.22/√V nanometre

lambda = 0.11 nanometre

Hence, the de Broglie wavelength of an electron is accelerated through a potential of 120V is 0.11 nanometre.

Answer 2

Answer:

0.1121 nm

Explanation:

• Given :- V = 120V
• To find :- de Broglie wavelength of electron(lambda) in nm
• Formula :- 1.228/√V
• Calculation :-

From formula,

lambda = 1.228/√120

Using log- antilog

= antilog { log (1.228)-0.5× log (120)}.

.°. [√ has ½ power in decimal 0.5]

= antilog {0.0892 - 0.5× 2.0792}

= antilog { 1.0496}

= 0.1121 nm

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